Prove it is a geometric progression.

Question

Question

Prove that if for an exponential function $ y = a^x (a>0;a \neq 1)$ the value of the argument $x=x_n(n=1,2,...)$ form an arithmetic progression, then the corresponding values of the function $y_n=a^{x_n}(n=1,2,...)$ form a geometric progression.

Answer 1

We know the following $$ y = a^x $$ Now $$y_1 = a^{x_1}$$ $$y_2 = a^{x_2}$$ $$y_3 = a^{x_3}$$

$$\frac{y_3}{y_2} = \frac{a^{x_3}}{a^{x_2}}$$ $$\frac{a^{x_3}}{a^{x_2}} = a^{x_3-x_2}$$ $$a^{x_3-x_2} = a^{(A+2d)-(A+d)}$$ $$a^{(A+2d)-(A+d)} = a^d$$

Similarily

$$\frac{y_2}{y_1} = a^d$$

Hence we can show that $$ \frac{y^3}{y^2} = \frac{y_2}{y_1}$$

Therfore, we have proven that it is a geometric progression.

Answer 2

Let an arithmetic sequence for exponents be $$ a, a+d, a+2d,....$$

The corresponding exponentials are $$ e^a ,e^{a+d}, e^{a+2d},....$$

$$e^a ,e^d e^a, e^{2d}e^a,....$$

$$e^a ,r e^a, r^2e^a,....... $$

Which is a geometric sequence with r= e^d.